Django models.FileField(upload_to=function1, default=function2)

Question

I'm trying to set the default image in a django model with a function. Te reason is that i generate the image (a qrcode) and i want to save it with a model asociated with the request.user, it is working with the upload_to= but fails with the default, it seems it doesn't get the instance.

The code is: -- models.py --

from django.db import models
from django.contrib.auth.models import User
import django.utils.timezone as timezone
from myproject.settings import MEDIA_ROOT
import os

def function1(instance, filename):
    retun os.path.join(self.uid.username, str(instance.shortname), '.jpg')

def function2(instance):
    return os.path.join(MEDIA_ROOT, str(instance.username), 'tmp.jpg')

class MyModel(models.Model):
    uid = models.ForeignKey(User)
    shortname = models.CharField()
    qr = models.FileField(upload_to=function1, default=function2)

this gives the function2() takes exactly 1 argument (0 given)

i also tried to pass uid as parameter to function2 like this:

def function2(uid):
    username = uid.username
    return os.path.join(MEDIA_ROOT, username, 'tmp.jpg')

class MyModel(models.Model):
    uid = models.ForeignKey(User)
    shortname = models.CharField()
    qr = models.FileField(upload_to=function1, default=function2(uid))

but this won't work too, gives an AttributeError exception: 'ForeignKey' object has no attribute 'username'

Any ideas on how can i tell django to upload a locally generated file (generated on the django machine at MEDIA_ROOT/request.user.username/)

Thanks!!

Why would you think that the default function would be passed the instance? It is not, and that's consistently the case. — Daniel Roseman
– Daniel Roseman, Commented Nov 9, 2015 at 20:20

Rafa He So · Accepted Answer · 2015-11-09 20:27:21Z

1

I found a way to solve my problem from views here: Loading Django FileField and ImageFields from the file system

answered Nov 9, 2015 at 20:27

Rafa He So

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2 Comments

Rafa He So Over a year ago

Thanks for your answers! will try them any way!

rhoerbe Over a year ago

This answer conveys the idea that you need to create a django.core.files.File object for FileField first.

Jared Mackey · Accepted Answer · 2015-11-09 20:27:03Z

0

You need to be using instance instead of self. Also, function2 will not be being passed an instance.

answered Nov 9, 2015 at 20:27

Jared Mackey

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Comments

Gocht · Accepted Answer · 2016-09-13 14:49:27Z

0

You need to use instance instead of self. self does not exist in function1 and function2 scope:

def function1(instance, filename):
    return os.path.join(instance.uid.username, str(instance.shortname), '.jpg')

def function2(instance):
    return os.path.join(MEDIA_ROOT, str(instance.username), 'tmp.jpg')

class MyModel(models.Model):
    uid = models.ForeignKey(User)
    shortname = models.CharField()
    qr = models.FileField(upload_to=function1, default=function2)

You can found docs for FileField.upload_to here

You get function2() takes exactly 1 argument, (0 given) because function2 is expecting for instance param, and if you try to pass uid like this:

class MyModel(models.Model):
    uid = models.ForeignKey(User)
    shortname = models.CharField()
    qr = models.FileField(upload_to=function1, default=function2(uid))

will raise an error because at this point uid is a models.ForeignKey instance. You could try to do this in a pre_save signal:

def function2(sender, instance, *args, **kwargs):

    if not instance.qr:
        instance.qr = os.path.join(MEDIA_ROOT, username, 'tmp.jpg')

pre_save.connect(function2, sender=MyModel)

edited Sep 13, 2016 at 14:49

answered Nov 9, 2015 at 20:26

Gocht

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1 Comment

Rafa He So Over a year ago

the instance vs self was an error while copying the code... i edited it on the first post to correct it Thanks!, tried this but it still says: function2() takes exactly 1 argument, (0 given)

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